BÀI TẬP VẬT LÝ - LỚP 10

39.2. Khi dp sudt ridng phdn cua hcri nudc trong khdng khf tang thi dp dm tuydt ddi cfla khdng khf tang, giam hay khdng ddi ? Tai sao ? A. Tang. Vi khi dp sudt rieng phdn cua hoi nudc trong khdng khi tang thi lupng hoi nudc cd trong 1 m khdng khf tang. B. Tang. Vi khi dp sud't rieng phdn eua hoi nudc trong khdng khf tang thi dpng nang ehuyin dpng nhidt cua cac phdn tfl hoi nude trong khdng khf tang. C. Khdng ddi. Vi khi dp sud't ridng phdn cua hoi nudc trong khdng khf tang 3 •> thi lupng hoi nudc cd trong 1 m khdng khi hau nhu khdng thay ddi. D. Giam. Vi khi dp sud't ridng phdn cua hoi nudtc trong khdng khf tang thi lupng hoi nudc cd trong 1 m khdng khf giam. 39.3. Khi nhidt dp khPng khf tang thi dp dm tuyet ddi va dp dm ti ddi cua nd thay ddi nhu thi nao ? A. Dp dm tuydt ddi va dp dm cue dai diu tang nhu nhau ndn dp dm ti ddi khdng thay ddi. B. Dp dm tuydt ddi giam, edn dp am cue dai tang nen dp dm ti ddi giam. C. Dp am tuydt ddi tang chdm, cdn dp am cue dai tang nhanh hon ndn dp dm ti ddi giam. D. D6 dm tuyet ddi khdng thay ddi, edn dp am cue dai giam ndn dp dm ti ddi tang. 39.4. Nudc ndng hon khdng khi. Tai sao trong cung dieu kidn nhidt dp va dp sudt, khdng khf khd lai nang hon khdng khf dm ? 39.5. Tai sao khi nhidt dp cfla khong khf dm tang ldn thi dp am tuyet ddi lai tang va dp dm ti ddi eua khdng khf lai giam ? 39.6. Tai sao trong nhiing ngay ndng ndng va dm udt, ta lai cam thdy khd chiu hon so vdi nhiing ngay ndng ndng nhung khd rao ? 39.7. Tai sao trong nhung ngay he ndng bfle thi vl ban ddm lai cd nhilu suong hon ? 39.8. Can cfl cac sd do dudi day cua tram quan sat khi tupng, hay cho bid't khong khf budi sang hay budi trua mang nhieu hoi nudc hon ? Giai thfch tai sao. - Budi sang : nhiet dp 20°C, dp dm ti ddi 85%. - Budi trua : nhidt dp 30°C, dp dm ti ddi 65%. - Khdi lupng ridng cua hoi nudc bao hoa d 20°C la 17,30 g/m^ va d 30°C la 30,29 g/ml 99

BAI TAP CUOI CHUUNG Vll VII.l. Khi ddt ndng mpt vanh kim loai ddng chat thi dudng kfnh ngoai va dudng kfnh trong cua nd tang hay giam ? A. Dudng kfnh ngoai va dudng kinh trong diu tang. B. Dudng kfnh ngoai va dudng kfnh trong diu giam. C. Dudng kinh ngoai tang va dudng kfnh trong giam. D. Dudng kfnh ngoai giam va dudng kfnh trong tang. VII.2. Khi vat rdn kim loai bi nung ndng thi khdi lupng ridng eua nd tdng hay giam ? Giai thfch tai sao. A. Tang. Vi the tfch cfla vat khdng ddi, nhung khdi lupng cua vat giam. B. Giam. Vi khdi lupng efla vat khdng ddi, nhung thi tfch cua vat tang. C. Tang. Vi thi tfch cua vat tang cham, cdn khdi lupng cfla vat tang nhanh hon. D. Giam. Vi khdi lupng cua vat tang eham, cdn thi tieh cua vat tang nhanh hon. VII.3. Tren tdm thep cd mpt IP thung hinh trdn. Khi bi nung ndng, didn tfch Id thflng thay ddi nhu the nao ? Nd'u didn tfch Id thflng d 0°C la 5 mm^ thi b 500°C se bdng bao nhieu ? Hd sd nd dai cua thep la 12.10\"^ K~\\ A. Giam. Dien tfeh Id thung d 500°C bdng 4,53 mm^. B. Tang. Didn tfch Id thflng d 500°C bdng 5,03 mm^. C. Tang. Didn tfch Id thung d 500°C bdng 5,06 mm^. D. Giam. Dien tfeh Id thung d 500°C bdng 4,92 mm^. VII.4. Dat mpt que diem ndi tren mat nude nguydn chd't. Nd'u nho nhe vai giot nude xa phdng xud'ng mat nudc gdn mpt eanh cua que diem thi que didm se dung yen hay chuyin ddng ? Vi sao ? A. Dung yen. Vi nudc xa phdng khdng lam thay ddi luc cang be mat cua nudc. 100

B. Chuyin dpng quay trdn. Vi nudc xa phdng hod tan trong nudc cd tac dung lam thay ddi luc cang bl mat cfla nudc, do dd se gdy ra cac luc lam quay que didm. C. Dich chuyen vl phfa nudc xa phdng. Vi nudc xa phdng cd he sd cang be mat ldn hon he sd cang bl mat cua nudc. D. Dich chuyen vl phfa nudc nguydn chdt. Vi nudc xa phdng cd he sd cang bl mat nhd hem so vdi nudc nguydn chd't. VII.5. Cd ndn dung nut bpe gie (bang vai spi bdng) dl nflt chat mieng chai dung ddy xang hoac ddu hoa khdng ? Vi sao ? A. Ndn dung nut bpe gie. Vi nut bpe gie mim, dd nut chat midng ehai ndn xang ddu trong chai khdng bi bay hoi ra ngoai. B. Khdng ndn dflng nut bpe gie. Vi xang ddu se thdm theo gie do tdc dung mao ddn cfla cae spi vai dl \"bd\" ddn ra ngoai midng chai va bay hoi. C. Khong nen dflng nut bpe gie. Vi nut bpe gie hay bi mun va dd chay. D. Ndn dflng nut bpe gie. Vi nflt bpe gie dl kiem va khdng bi xang ddu thdrn udt. VII.6. Phai lam theo each nao sau ddy dl tang dp cao cua cpt nudc trong dng mao ddn? A. Giam nhidt dp cfla nudc. B. Dflng d'ng mao ddn ed dudng kfnh ldn hon. C. Pha them rupu vao nudc. D. Dflng d'ng mao ddn cd dudng kfnh nhd hon. VII.7. Hay ndu rd each phan loai cdc chdt rdn va nhung dac tfnh cua mdi loai chd't rdn dd. VII.8. Tai sao tinh thi kim cuong lai bin viing hon tinh the than chi, mac du chung diu cd'u tao tfl cflng eae nguydn tfl cacbon ? VII.9. Hay ndu rd each phdn loai bid'n dang eo cua vdt rdn va su khdc nhau giua cdc loai bie'n dang dd. Phdt bilu va vid't bilu thflc cua dinh ludt Hue vl bid'n dang CO cfla vdt rdn. Nhung loai bid'n dang co ndo tudn theo dinh ludt nay ? VII.IO. Mpt thanh rdn AB ddng chat, tilt didn diu cd khdi lupng m = 100 kg dupc treo bdng ba spi ddy thdng dung va each diu nhau (H.VII.l). 6 chfnh gifla 101

thanh rdn la spi ddy thep, edn d hai ddu yyyy/y//yy//yyy'yyy/y. thanh rdn la hai spi ddy ddng. D6 dai va tilt dien cua ba spi ddy gid'ng nhau. Sud't Day d6ng dan hdi cfla thep ldn gdp 1,2 ldn cfla ddng. Hay tfnh luc eang efla mdi spi day - Day thep sao cho thanh rdn luon ndm ngang. Ldy 3 1 sl g = 9,8 m/sl Day dong Hinh VII.l VII.l 1. Mpt thanh sdt trdn cd dp ddi ban ddu /g - 50 em va tilt didn ngang S = 2,5 mm . Keo dan thanh sat bang mdt luc F cd dp ldn tang ddn va do dp dan dai A/ tUdng flng cfla thanh sdt. Kit qua cua cae ldn do dupc ghi trong bang sd lieu dudi ddy : F(N) A/ (mm) a = - (N/m^) A/ 8= 100 0,10 200 0,20 k 300 0,30 400 0,40 500 0,50 600 0,60 a) Tfnh dp dan dai ti ddi £ cua thanh sdt va flng sud't a cfla luc keo tdc dung ldn thanh sdt trong mdi ldn do. b) Ve dd thi bilu didn su phu thupc cua dp dan dai ti ddi s vdoflngsud't a. c) Dua vao dd thi ve dupe, tim gid tri eua sud't dan hdi E va he sd dan hdi k. VII.l2. Hay phdn bidt su nd dai va sir nd khdi efla vdt rdn. Phdt bilu va viet cdng thflc xdc dinh dp nd dai va dp nd khdi cfla vat rdn. Cac cdng thflc nay chi dp dung cho nhung vat rdn cd dac tinh nhu the nao ? Vll. 13. Mpt thanh thep b 20°C ed tilt didn 4 cm^ va hai ddu cfla nd gdn chat vao hai bfle tudng ddi dien. Tinh luc do thanh thep tae dung ldn hai bfle tudng nd'u nd bi nung ndng dd'n 200°C. Thep cd sud't dan hdi cua E = 20.10'° Pa va cd hdsdndddi a = 12.10\"^ K\"'. 102

VII.14. Khi tie'n hanh thf nghiem khao sat t(°C) IQ - 500 m m A/ su nd dai vi nhiet efla vdt rdn, cdc A/(mm) lo . kit qua do dp dai IQ cua thanh thep 20 0,12 d 0°C va dp nd dai A/ cfla nd ung 30 0,18 vdi dp tang nhidt dp t (tfnh tfl 0°C 40 0,24 dd'n t°C) dupc ghi trong bang 50 0,30 sd lidu bdn. 60 0,36 70 0,42 a) Tfnh dd dan dai ti ddi — cfla 80 0,48 lo thanh thep d nhiing nhidt dp t khdc nhau trong bang sd lieu. b) Ve dd thi bilu didn su phu thupc cua do dan ddi ti ddi — vao nhidt dp t efla thanh thep. c) Dua vao dd thi ve dupe trong edu trdn, tfnh gia tri trung binh cfla he sd nd dai a cfla thep. VII.l5. Mpt dng nhd gipt dung thang dung ben trong dung nudc. Nude dfnh udt hoan todn midng d'ng va dudng kfnh mieng d'ng la 0,43 mm. Trpng lupng mdi gipt nude roi khdi midng dng la 9,72.10\" N. Tfnh he sd cdng bl mat cfla nudc. VII.16. Mpt vdng nhdm cd trpng lupng la 62,8.10 N dupc ddt sao eho day cua nd tid'p xflc vdi mat chdt long dung trong mpt cdc thuy tinh. Dudng kfnh trong va dudng kfnh ngoai cua vdng nhdm ldn lupt bdng 48 mm va 50 mm. Tfnh luc keo vdng nhdm dl biit nd len khdi mat thoang cfla ehd't long trong hai trudng hdp: a) Chdt ldng la nude. b) Chd't long la rupu. He sd cang bl mat cua nudc la 72.10\"^ N/m va cfla rupu la 22.10~^ N/m. VII.l7. Tai sao nhidt dp cua mat nudc hd hoac ao vl mfla he lai thdp hon nhidt dp cua khdng khf d phfa trdn mat nudc ? VII.18. Cd thi lam cho nudc sdi bdng each lam lanh nd dupc khdng ? Tai sao ? 103

Vll. 19. Tfnh nhidt lupng cdn cung cd'p dl bid'n ddi 6,0 kg nude dd d -20°C thanh hoi nudc d 100°C. Nhidt ndng ehay ridng efla nudc da la 3,4.10^ J/kg. Nhidt dung rieng cfla nudc da la 2 090 J/(kg.K). Nhidt dung ridng cua nudc la 4 180 J/(kg.K). Nhidt hod hoi ridng cua nudc la 2,3.10^ J/kg. VII.20. Ngudi ta dd 0,20 kg chi ndng ehay d 327°C vao mpt cdc chfla 0,80 / nudc d 15°C. Trong qua trinh nay da cd 1,0 g nudc bi bid'n thanh hdi nudc. Tinh nhiet dp efla nudc cdn lai trong cdc d trang thai edn bdng nhidt. Nhiet ndng chay rieng cfla chi la 2,5.10 J/kg va nhiet dung rieng efla ehi la 120 J/(kg.K). Nhidt dung rieng cfla nudc la 4 180 J/(kg.K) va nhiet hod hcri ridng cfla nudc la 2,3.10 J/kg. Bd qua su md't mat nhidt do cdc hdp thu va do nhidt truyin ra ngoai cdc. VII.21. Tai sao trong nhung ngay mfla ddng gia lanh, ta lai cd thi nhin thdy hoi thd cfla chfnh minh va cdc mat kinh cfla sd lai dd bi \"dd md hdi\" khi trong phdng cd ddng ngudi ? VI1.22. Nhidt dp eua khong khi trong phdng la 20°C. Nd'u cho may dilu hod nhiet dp chay dl lam lanh khdng khf trong phdng xudng tdi 12°C thi hoi nudc trong khdng khf cfla can phdng trd ndn bao hod va tu lai thanh suong. Nhidt dp 12°C dupc gpi la \"dilm suong\" cua khong khf trong can phdng. Hay tfnh dp dm tuydt ddi va dp am ti ddi cua khong khf trong cdn phdng nay. Kfeh thudc efla can phdng la 6x4x5 m. Khdi lupng ridng cfla hcri nudc bao hod trong khdng khf d 12°C la 10,76 g/m^ va d 20°C la 17,30 g/ml 104

i - m dill - 4ll/ffH(i MN Clfll - MP 50 ChitfrngI DONG HOC CHAT DIEM BAI 1 1.1. l ^ d ; 2 ^ d ; 3 ^ g ; 4 - ^ e ; 5 - ^ b ; 6 ^ c ; 7 - > ' a . 1.2. D ; 1.3. C ; 1.4. B ; 1.5. D ; 1.6. D ; 1.7. C. t:^vi:y,::i;vi-ii-Jii^J-iviv:: 1.8*. a) Khi xudng chay xudi theo ddng thi quy dao : 0 - - - .-..\"...\"..\"__ -— eua nd se la mpt dudng thdng song song vdi bd sdng, do dd ndn ehpn (H. 1. IGa): 3 - Mpt vat mde gdn cd dinh vdi bd sdng tai vi tri a) ^' xudt phdt O trdn bin sdng. y - Mpt true toa dp Ox ndm dpe bd sdng va hudng ^^M: : theo ehilu ddng chay. O: 7 ^ b) Khi xudng ehay vudng gdc vdi ddng chay thi b) quy dao cfla nd la mdt dudng thdng xidn gdc vdi bd sdng, do dd nen chpn (H.1.1 Gb): \"'\"'' ^'^^ - Mpt vdt mdc gdn cd dinh vdi bd sdng tai vi tri xudt phat O trdn bin sdng. - Hai true toa dp vuong gdc Ox va Oy : True Ox ndm dpe bd sdng hudng theo chiiu ddng ehay, true Oy ndm vudng gde vdi bd sdng tai vi trf xud't phat O va hudng theo chiiu tfl vi trf xudt phdt O tdi vi trf ddi dien vdi nd d bd bdn kia. 1.9. Trudng hpp nay quy dao efla d td triing vdi qud'c Id 5, do dd nen chpn : - Mpt vdt mdc gdn cd dinh vdi bd'n xe Ha Npi tai vi tri xudt phat O. - Mpt true toa dp cong cd gd'c tai vi tri xudt phdt O, cd dang triing vdi qude Id 5 va hudng theo chiiu duong tfl Ha Npi tdi Hai Phdng. 1.10. a) Ddi vdi hanh khach len xe tai Ha Npi thi bin xe Ha Npi dupc chpn lam mde dudng di va thdi dilm xe d td bdt ddu xud't phat dupc chpn lam mdc thdi gian. Trudng hpp nay, khoang thdi gian dl xe d td chay tdi Hai Phdng la : 8 gid 50 phut - 6 gid = 2 gid 50 phut. 105

va quang dudng xe d td da chay dl tdi Hai Phdng dung bdng dp dai cua doan dudng Ha Npi - Hai Phdng, tu'c la bdng 105 km. b) Ddi vdi hanh khach ldn xe tai Hai Duong thi bd'n xe tai Hai Dudng dupc chpn lam mdc dudng di va thdi diem xe d td tid'p tue chay tfl Hai Duong dupe ehpn lam mdc thdi gian. Trudng hpp nay, khoang thdi gian di xe d td chay tdi Hai Phdng la : 8 gid 50 phut - (7 gid 15 phflt + 10 phflt) = 1 gid 25 phut va quang dudng xe d td da chay tfl Hai Duong tdi Hai Phdng la : 105 km - 60 km = 45 km BAI 2 2.1. l ^ c ; 2 ^ ^ g ; 3 ^ d ; 4 - » a ; 5 - » b ; 6 ^ d . 2.2. D ; 2.3. B ; 2.4. D ; 2.5. D ; 2.6. A ; 2.7. A ; 2.8. C ; 2.9. D ; 2.10. C. 2.11. May bay phai bay lidn tuc trong khoang thdi gian bdng : t = - = ^ ^ = 2,6gid = 2gid36phut V 2500 BF 2.12. a) Ngudi ldi xe phai cho d td chay lidn tuc vdi vdn td'c bdng : s V= - t Thay s = 120 km va t = 8 gid 30 phut - 6 gid = 2 gid 30 phut = 2,5 gid, ta tim dupe : V = — = 48 km/h 2,5 b) Khoang thdi gian dl P td chay ngupe tfl dia dilm B vl tdi dia dilm A bdng : t', = —s = 120 = 2^ gi.d. v' 60 Nhu vay d td chay vl tdi dia dilm A vao luc : (8 gid 30 phflt + 30 phut) + 2 gid = 11 gid 106

2.13. Khoang thdi gian am truyin trong khdng khf bdng : t = -s = 200 ~ 0„,5^9„ s V 340 Nhu vay khoang thdi gian chuyin dpng cfla vidn dan trong khdng khf bdng : t'= 1,0-0,59 = 0,41 s Tfl dd suy ra van tdc cua vien dan B40 chuyin dpng trong khdng khf bdng : 200 « 488 m/s t' 0,41 2.14*. a) Xe II xudt phdt luc 1 gid. b) Quang dudng AB dai 60 km. e) Vdn tdc cua xe I : Vj = 2 4 km/h ; cua xe I I : Vjj = 30 km/h. 2.15. a) Cdng thflc tfnh quang dudng di dupc va x(km) phuong trinh ehuyin dpng : 160 M - cfla xe may xud't phdt tfl A lfle 6 gid : 120 In Sj = vit = 40t 80 Xj = Sl = 40t vdi XQ = 0 - cfla d td xud't phdt tfl B Iflc 8 gid : S2 = V2(t - 2) = 80(t - 2) vdi t > 2 40 X2=Xo + S2=20 + 8 0 ( t - 2 ) b) Dd thi toa dp cfla xe mdy va d td dupe 0 ve trdn hinh 2.1G. Dudng I la dd thi eua xe may. Dudng II la dd thi cua d td. t(h) Hinh 2.1G c) Trdn dd thi hinh 2.1G, vi tri va thdi dilm d td dudi kip xe mdy dupe bilu diln bdi giao dilm M cd toa dp : X]yj = 140 km tf^ = 3,5 h d) Kilm tra lai kit qua tim dupe bang each giai phuong trtnh : X, = X2 =^ 40t = 20 + 80(t - 2) 6 td dudi kip xe may sau thdi gian : _ 140 ' ^ \" l o = 3,5h Thdi dilm d td dudi kip xe may la : 6 + 3,5 = ^,5 h 107

va vi trf d td dudi kip xe may : XM = 40.3,5 = 140 km. 2.16*. Gia sfl ngudi do gap d td tai dilm N. a) A H ^i E Khoang thdi gian t de ngudi dd chay i^. A tfl M tdi N phai dflng bdng khoang thdi ~~~L--,, V ' ' ' gian dl d td chay tfl A tdi N (Hai trudng hpp, hinh 2.2G). M Ta cd AN = Vjt = 36t (AN do bdng NH B kildmet va t do bdng gid). M MN = V2t=12t Hinh 2.20 AH = VL^-h^ = 0,19365 km HN= VMN^-h^= Vl2V^a05^ Ca 2 trudng hpp, diu cd : HN^ = MN^ - h2 Cudi cung ta dupe phuong trinh bdc hai: 1152t^ - 13,9428t + 0,04 = 0. t = 0,00743 h = 26,7 s AN = 0,26748 km Giai ra ta dupc hai nghiem : AN = 0,16812 km t = 0,00467 h = 16,8 s. Quang dudng MN ma ngudi d'y phai chay la MN = 89,2 m hoac MN = 56 m. Gpi a la gdc tao bdi MN va MH : cosa, = = « 0,5605 ; a, «55°54' MN 89,2 ' cosa, = = — « 0,8928 ; a , « 26°46' ^ MN 56 ^ Kit qua If thu nhung rdt khd giai thfch la ed hai ddp sd diu cd thi chap nhdn dupc. 2.17*. Gpi tl la khoang thdi gian d td di dupc doan dudng Si vdi tdc dp Vi va t2la khoang thdi gian d td di dupc doan dudng S2 vdi td'c dp V2. Tdc dp trung binh cfla P td dupc tfnh theo cdng thflc : S Si + ST t tl + t2 108

Vi tl = tj = - , nen Si = Viti = Vi - va S2 = V2t2 = V2- . Thay cac gia tri nay .Z L ^ vao (l),ta tim dupc : tt (2) V, h Vo — „ , „ v*=^f^ =^ '—+— TTTh,ay so\", .ta co' : v,i, = 60 +2 402= 5.„0 ,km/-h 2.18.* Tde dp trung binh efla xe dap dupc tfnh theo cong thflc (1). Vdi Sl = S2 = - , ta cd tl = — = va t2 = — = . Thay cac gid tri 2 Vj 2vi V2 2v2 nay vdo (1), ta tim dupc : ss V - 2 \" ^ 2 _ 2viV2 ^tb _ ! _ + _ ^ Vl + V2 2vi 2v2 Tfnh bdng sd : v.u = ——'-— = 14,4 km/h. ' \" 1 2 + 18 Ghi chu : Khdng thi tfnh td'c dp trung binh bdng gid tri trung binh cpng cua cdc tde dp chuyen dpng tren nhiing doan dudng k l tid'p, trfl khi cac khoang thdi gian ehuyin dpng vdi cdc td'c dp khae nhau trdn nhirng doan dudng ke tie'p deu bdng nhau. Vf du nhu trudng hpp neu trong bai tap trudc hoac trudng hpp chuyin dpng thdng bid'n ddi diu thi ta cd thi tfnh : _ Vl + V2 ^tb - ~ BAIS 3.1. l - ^ e ; 2 ^ d ; 3 - ^ n ; 4 ^ i ; 5 - > / ; 6 - > k ; 7 ^ m ; 8 - ^ d ; 9 ^ b ; l O ^ a ; l l ^ h ; 12-»g; 13-^e. 3.2. A ; 3.3. D ; 3.4. A ; 3.5. D ; 3.6. A ; 3.7. C ; 3.8. B ; 3.9. D ; 3.10. A. 109

3.11*. Chpn true toa dp trflng vdi dudng thdng AB va ehilu dirong hudng tfl A dd'n B. a) Hai d td chay cflng chiiu (H.3.1G): 6 td xudt phdt tfl A chay theo chiiu duong (+) va ehuyin dpng nhanh ddn ndn gia tdc aj cfla nd cflng chiiu vdi vdn td'c Vl tflc la hudng theo ehilu ehuyin ddng tfl A dd'n B. Cdn d td xudt phdt tfl B cung chay theo ehilu duong (+) va chuyin dpng chdm ddn ndn gia tdc a2 cfla nd ngupc chiiu vdi vdn td'c V2 (v2 cung hudng tfl A den B). Trong trudng hpp nay, gia td'c aj va a2 eua hai d td ngupe hudng (cflng phuong, ngupc ehilu). [^^£)^^^i '^-{^zlZh-v. Hinh 3.1G b) Hai P td ehay ngupc ehilu (H.3.2G) : 6 td xud't phdt tfl A chay theo chieu duong (+) va chuyin dpng nhanh ddn ndn gia td'c ai eua nd cflng ehilu van td'c Vl, tflc la hudng tfl A dd'n B. Cdn d td xudt phdt tfl B chay ngupc chiiu duong (+) va chuyin dpng chdm ddn nen gia tde a2 eua nd ngupc chiiu van tde V2, tfle la cung hudng tfl A dd'n B. Trong trudng hop nay, gia tdc ai va a2 cflng hudng (eiing phuong, cung chiiu). ^^-Q=^^^i Hinh 3.2G 3.12. Can cfl vao dd thi vdn tdc eua 4 vdt I, II, III, IV ve tren hinh 3.2, ta cd thi xac dinh dupc van tdc ddu VQ va van td'c tfle thdi v cua mdi vat chuyen dpng, do dd tfnh dupc gia tdc theo edng thflc : a= Sau dd thay cdc gia tri tim dupc vao cdng thflc tfnh van td'c v va edng thflc tfnh quang dudng di dupc cfla mdi vdt ehuyin dpng : V = VQ + at at2 110

2 - V a t I : v o = 0 ; v = 2 0 m / s ; t = 2 0 s ; a = — = 1 m/s^ ; v = t; s= —• 20 2 - vat II: VQ = 20 m/s ; v= 40 m/s ; t = 20 s ; a = — = 1 m/s ; t^ v = 20 + t ; s = 20t+ — 2 - vat III: v= VQ = 20 m/s ; t = 20 s ; a = 0 ; s= 20t. 40 2 - vat IV : VQ = 40 m/s ; v = 0 ; t = 20 s ; a = = - 2 m/s^ ; V = 40 - 2t; s = 40t - t l 3.13. a) Chpn true toa dp trung vdi quy dao ehuyin dpng thang cfla P td, ehilu duong cfla true hudng theo chiiu ehuyin dpng. Chpn mde thdi gian la luc d td bdt ddu tang ga. Gia tdc cua d td bdng : a = ^l^^ t Thay sd : a= = 0,2 m/s . b) Vdn tde eua d td sau 30 s kl tfl khi tang ga : V = VQ + at Thay sd : v = 12 + 0,2.30 = 18 m/s. e) Quang dudng d td di dupc sau 30 s kl tfl khi tang ga : at2 S=VQt-.— Thay sd : s = 12.30 + ^'^'^^^^ = 450 m. 3.14. a) O tP dang chuyin dpng vdi vdn td'c VQ = 36 km/h = 10 m/s thi xud'ng dd'c va chuyin dpng thdng nhanh dan diu vdi gia td'c a = 0,2 m/s . Do dd quang dudng d td di dupc trong khoang thdi gian t dupc tfnh theo cPng thfle : at2 S = VQt+ — Thay sd, ta ed : 9 6 0 = 1 0 t + ^ - — hay t^+100t - 9 600 = 0 111

Phuong trinh trdn cd hai nghiem sd : ti = 60 s va t2 = - 1 6 0 s. 6 ddy chi lay nghiem sd duong t, = 60 s va loai nghiem sd am t2 = - 1 6 0 s. b) Van tdc d td d cudi doan dde tfnh theo cdng thflc : V = VQ + at Thay sd, ta tim dupc : V = 10 + 0,2.60 = 22 m/s = 79,2 km/h. 3.15. Trong chuyen dpng thang nhanh ddn ddu, vdn tdc lien he vdi quang dudng di dupc va gia tdc theo cdng thflc : 92 v\" - VQ = 2as Gpi Vl la van td'c cua doan tau sau khi ehay dupc doan dudng Si = 1,5 km va V2 la van td'c cfla doan tau sau khi chay dupc doan dudng S2 = 3 km k l tfl khi doan tau bdt ddu rdi ga. Vi gia td'c a khdng ddi va vdn tdc ddu VQ = 0, ndn ta ed : 22 VI = 2 a s i ; V2 = 2as2 Tfl dd suy ra : ^2 ^2 vf Sl Thay sd, ta ed : V2 = 36 1 — = 50,91 km/h ^ 51 km/h. ^ 1, J 3.16*. Trong chuyin dpng thdng nhanh ddn diu khdng vdn tde ddu, quang dudng vidn bi di dupc sau khoang thdi gian t lidn he vdi gia td'c a theo cdng thflc : s-^ 2 Nhu vay quang dudng vien bi di dupc sau khoang thdi gian t = 4 s la : *2 va quang dudng vien bi di dupe sau khoang thdi gian t = 5 s la : S5 = a(5)2 = 1102,^5a 112

Tfl dd suy ra quang dudng vidn bi di dupc trong gidy thfl nam la : As = S5 - S4 = 12,5a - 8a = 4,5a Theo ddu bdi: As = 36 cm, ndn gia tdc cfla vidn bi la : 36 = 4,5a =^ a = = 8 cm/s 4,5 b) Theo ket qua trdn, ta tim dupc quang dudng vidn bi di dupe sau khoang thdi gian t = 5 s la : S5 = 12,5.8 = 100 em. 3.17.* a) Trong chuyin dpng thang nhanh dan diu vdi van td'c ddu VQ, quang dudng vdt di dupe sau khoang thdi gian t lidn hd vdi gia td'c a theo cdng thiic : at2 S = Vot+ — Nhu vdy quang dudng vdt di dupe sau khoang thdi gian t = 4 s la : A a(4)2 . 54 = V o - 4 + - ^ — = 4 v Q + 8a va quang dudng vdt di dupc sau khoang thdi gian t = 5 s la : c a(5)^ , 55 = VQ.5+ =5VQ+12,5a Tfl dd suy ra quang dudng vdt di dupc trong giay thfl nam la : As = Sg - S4 = (5VQ + 12,5a) - (4vo + 8a ) = VQ + 4,5a Theo ddu bai : Vo= 18 km/h = 5 m/s va As = 5,90 m, ndn gia tde cfla vien bi bang : AS-VQ 5,9-5 . . .2 a = = = 0,2 m/s 4,5 4,5 b) Theo kit qua tren, ta tim dupe quang dudng vdt di dupc sau khoang thdi gian t = 10 s bdng : 0 2 (10)^ SiQ = 5.10 + ' -^ ' = 50 + 10 = 60 m. 3.18.* a) Chpn true toa dp trung vdi quy dao chuyin dpng thdng cua 0 to, chiiu duong cua true hudng theo ehilu chuyen ddng. Chpn mde thdi gian la luc 8.BTVATLI10-A 113

d td bdt ddu ham phanh. Theo cdng thflc lidn hd gifla quang dudng di dupc vdi vdn tdc va gia td'c trong chuyin dpng thang cham ddn diu : 2 2o V - VQ = 2as ta suy ra gia td'c cfla d td bdng : 2s 2.125 Ddu - cfla gia td'c a ehiing td d td chuyin dpng thdng chdm ddn diu cd chiiu ngupc vdi ehilu duong da chpn trln true toa dp, tflc la ngupc chiiu vdi vdn tdc ban ddu VQ. b) Quang dudng d td di dupc trong chuyin dpng thang chdm ddn diu tinh theo cdng thflc : .2 at S = VQt + — Thay sd ta cd \"2 125=15.-«-5'' 2 hay t^-60t +500 = 0 Phuong trinh trdn cd hai nghidm : ti = 50 s va t2 = 10 s. Trong hai nghidm nay, ta phai loai ti = 50 s vi gid tri nay ldn hdn khoang thdi gian dl d td dflng lai (v = 0) kl tfl khi bdt ddu ham phanh. Thue vdy, khi d td diing lai thi ta cd dilu kien : V = VQ + at = 0 suy ra : t = - ^ = = 30 s < ti = 50 s a -0,5 ^ Nhu vdy khoang thdi gian dl d td chay thdm dupc 125 m ke tfl khi bdt ddu ham phanh la t2 = 10 s. 3.19.* a) Phucmg trinh chuyin dpng cua xe may xud't phat tfl A chuyin dpng nhanh ^^ ~2 2 ddn ddu khPng van tde ddu vdi gia tde ai = 2,5.10 m/s : Xi = ^ = ^'^-^\"\"'^' = l,25.10-¥ '2 2 1 1 4 S.BTVATLIIO-B

Phuong trinh chuyin dpng cua xe mdy xud't phdt tfl B cdch A mpt doan XQ = 400 m chuyin dpng nhanh ddn diu khong van tdc ddu vdi gia tde a2 = 2,0.10 ^m/s^ X2 = XQ + -_t ^ ==4.4^00«0++2M,0.1i 0t ^O*^l = 400 + 1,0.10\"¥ 22 b) Khi hai xe mdy gap nhau thi Xj = X2, nghia la : 1,25.10\"¥ = 400 + 1,0.10\"V hay 0,25.10\"¥ = 400 PhUdng trinh trdn cd hai nghiem : t = ± 400 s. O ddy chi giu lai nghidm duong t = + 400 s. Nhu vdy thdi dilm hai xe dudi kip nhau la 6 phflt 40 gidy kl tfl luc xud't phdt. Thay vao phucmg trinh trdn, ta tim dupc vi tri hai xe may dudi kip nhau la : Xl = l,25.10\"l(400)^ = 2.10^ m = 2 km. e) Tai vi tri hai xe mdy dudi kip nhau, xe xud't phdt tfl A cd van tdc bdng : Vl = ait = 2,5.10~l400 = 10 m/s = 36 km/h. cdn xe xudt phdt tfl B cd van td'c bdng : V2 = ajt = 2,0.10~l400 = 8 m/s = 28,8 km/h. BAI 4 4.1. l ^ d ; 2 - > d ; 3 ^ e ; 4 - > b ; 5 - > a ; 6 ^ c . 4.2. C ; 4.3. B ; 4.4. C ; 4.5. D ; 4.6. A ; 4.7. B ; 4.8. C ; 4.9. C. 4.10. Nd'u gpi s la quang dudng vidn da di dupe sau khoang thdi gian t kl tfl khi bdt ddu roi tdi khi cham dd't va gpi Si la quang dudng vidn da di dupc trudc khi cham ddt 1 s, tflc la sau khoang thdi gian t^ = t -1 thi ta cd cac cdng thflc : 2 _ g(t -1)^ ^1 ~ '^ 115

Tfl dd suy ra quang dudng vidn da di duoc trong 1 s cudi trudc khi cham dd't la: As = s - s i = g ^ - g ( ^ - ^ ) ' = g t - g ' 2 2 ^2 Vdi As = 24,5 m va g = 9,8 m/s , ta tim dupc khoang thdi gian roi tu do cua vidn da: As 1 24,5 ^ , , t = — + - = —^ + 0,5 = 3 s. g 2 9,8 4.11. Quang dudng ma vdt roi tu do di dupc sau khoang thdi gian t tfnh theo cdng thflc : 2 Tfl dd suy ra quang dudng md vdt roi ttr do di dupc sau khoang thdi gian t = 3 s la : S 3 = ^ = 4,5g va quang dudng ma vdt roi tu do di dude sau khoang thdi gian t = 4 s la : . = ^ • = 3, • Nhu vdy quang dudng ma vdt roi tu do di dupc trong gidy thfl tu la : As = S4 - S3 = 8g - 4,5g = 3,5g = 3,5.9,8 = 34,3 m Vdn tdc cfla vdt roi tu do tfnh theo cdng thiic : v = gt Tfl dd suy ra, trong gidy thfl tu, van tdc cua vat da tang ldn mpt lupng bdng : Av = V4 - V3 = 4g - 3g = g = 9,8 m/s. 4.12. Chpn thdi dilm vidn bi A bdt ddu roi lam mdc thdi gian. Nd'u gpi t la thdi gian roi cfla vidn bi A thi thdi gian roi cua vien bi B se la t' = t + 0,5. Nhu vdy quang dudng ma vidn bi A va B da di dudc tfnh theo edc cdng thiic : A2 _ gt'2 _ g(t + 0,5)^ ' ^ - ^ - ~2 116

Tfl dd suy ra khoang cdch gifla hai vidn bi sau khoang thdi giah 2 s kitit khi bi A bdt ddu roi bdng : g(t + a 5 ) ^ gt^ _ g As = S B - S A = 2~~ 2 Tinh bdng sd: As = — (2 + 0,25) « 11 m. 2 4.13. Nd'u gpi s la quang dudng ma vdt da roi trong khoang thdi gian t va Si la quang dudng ma vat da rdi trong khoang thdi gian t' = t - 2 thi ta ed thi vid't: 2 ^ ^gt'^ ^g(t-2)^ ^2 2 Tfl dd suy ra quang dudng ma vat da di dupc trong 2 s cudi cung se bdng : ^ gt^ g(t - 2)^ As = s - Si = = 2g(t - 1) ^2 2 s 1 eat _et,2 Vdi As = - = - ^ - = ^ - , ta tim duoc : 4 42 8 ^ = 2g(t - 1) ^ t^ - 16t +16 = 0 8 Phuong trinh trdn ed hai nghidm sd : tj » 14,9 va tj ~ 1,07. CJ day ta chi gifl lai nghidm sd ti « 14,93 s va loai bd nghidm sd tj- 1,07 s < 2 s. Nhu vay, khoang thdi gian roi cfla vat ti « 14,9 s. Vdi tl « 14,9 s, ta tfnh dupc dp cao tfl dd vat rcri xud'ng ddt: 9.8.(14,9)^ , _^^ s = ^ 1 088 m. 2 4.14.* a) Trong trudng hpp thfl nhd't (khf cdu dung ydn) thi quang dudng vat roi tu do tfl dd cao s tinh theo cdng thflc : 2 117

Tfl dd suy ra khoang thdi gian rci tu do cua vat bdng : /2s /2.300 ^ „ t = J— = , «7,8s Vg V9,8 b) Trong trudng hop thfl hai (khf cdu dang ha xudng) thi vat roi nhanh ddn diu vdi vdn td'c ddu VQ = 4,9 m/s bdng vdn td'c ha xudng cua khf edu tfl dp cao s tfnh theo cong thflc : S = VQt + ^ \"2 Thay sd, ta thu dupc phuong trinh bdc hai nhu sau : 300 = 4 , 9 t + 9^ ^8^t^^ 2 hay t^ + t - M = o 4,9 PhUdng trinh nay cd hai nghidm sd va ta chi gifl lai nghidm sd duong : tj w 7,3. Nhu vay khoang thdi gian roi cua vat la tj » 7,3 s. c) Trong trudng hpp thfl ba (khi cdu dang bay ldn) thi lue ddu vdt dupc nem ldn cao vdi vdn tdc ddu VQ = 4,9 m/s bdng vdn tdc bay ldn cua khf cdu tfl dp cao s va chuyin dpng thang chdm ddn diu trong khoang thdi gian t2 ldn tdi dp cao ldn nhd't, tai dd vdn tdc v = 0. Khoang thdi gian t2 tfnh theo cdng thflc : V = VQ - gt, = 0 =}> to =Vn- ^ =4 -9^— = 0,5 s 0 62 2 g 9,8 Sau dd vdt lai rcri tu do tfl dp cao ldn nhdt xudng dd'n dp cao 300 m trong thdi gian t2 = 0,5 s, rdi tie'p tuc rdi nhanh ddn diu vdi van tde VQ = 4,9 m/s tfl dp cao 300 m xudng tdi dd't trong khoang thdi gian ti « 7,3 s (gid'ng nhu trudng hpp trdn). Nhu vdy, khoang thdi gian ehuyen ddng cfla vdt se bdng : t = 2t2 + tl = 2.0,5 + 7,3 = 8,3 s. BAIS 5.1.1->g;2->e;3^h;4^d;5^i;6^b;7->a;8^e;9->d. 5.2. D ; 5.3. C ; 5.4. C ; 5.5. D ; 5.6. C ; 5.7. A ; 5.8. B ; 5.9. D. 118

5.10. Tdc dp gdc cua diem A va dilm B bdng nhau : Tde dp dai cfla dilm A va dilm B khdc nhau : Vg corg rg Gia tdc hudng tdm cua diem A va dilm B khdc nhau : ^ ^ V A / T A ^ rfiVA ^ 1(2)2 = 2. ^B v ^ / r g r ^ v l 2 5.11. Tde dp gdc co va gia td'c hudng tdm ai,t cfla mpt dilm trdn vanh ngoai cua bdnh xe cd ban kfnh r = 25 cm = 0,25 m khi d td dang ehay vdi td'c dp dai V = 36 km/h = 10 m/s bdng : CO =V— = 10 =A4f^0 rAald/s r 0,25 r 0,25 5.12. Chu ki quay cfla Mat Trdng quay quanh Trai Ddt bdng : T = 27 (ngay-ddm) = 27.24.3 600 « 2,33.10^ s Td'c dp gdc cfla Mat Trang quay quanh Trai Dd't bdng : CO = 2—n « 2.3,1—4 - «^2^,7.,1^0_6° raJd,/s T 2,33.10^ 5.13. Td'c dp dai eua ddu kim phut va kim gid dupe tinh theo cPng thflc : 27trj Vl = corj = 2^2 V2 = cor2 = Tfl dd suy ra V2 r2 Tl 119

Thay ri = 1,5r2 ; Ti = 3 600 s ; T2 = 43 200 s vao cdng thflc trdn, ta tim dupc : Vl ^1.5r2 43200 _ ^ g V2 r2 3600 5.14. Tdc dp gdc va gia td'c hudng tdm cua vd tinh dupc tfnh theo cae cdng thflc : 271 2.3,14 3 CO = — = ~ 1,19.10 rad/s T 88.60 a^t = co^(R + h) = (1,19.10^^)^.6 650.10^ = 9,42 m / s l BAI6 6.1. l - > d ; 2 ^ d ; 3 ^ a ; 4 ^ e ; 5 ^ e ; 6 - > b . 6.2. D ; 6.3. C ; 6.4. B ; 6.5. B ; 6.6. B. 6.7. Gpi Vi 2 la vdn tdc eua d td (1) di tfl bin A ddi vdi d td (2) di tfl bin B, Vj 3 la van td'c cua d td (1) di tfl bin A ddi vdi bin xe (3) va V23 la van tdc cua d td (2) di tfl bin B ddi vdi bd'n xe (3). - Khi hai d td chay ngupe ehilu nhau thi P td tfl A tiln gdn lai B, ndn Vi 3 va Vl 2 eung phudng ehilu, edn V23 ngupe ehilu vdi Vi 3 va Vi 2. Do dd, theo cdng thflc cpng van tdc ta cd : Vl,3 = Vi,2-V2,3 suy ra : Vl,2 = Vl,3 + V23 O td (1) each d td (2) mpt doan dudng s = 20 km va ehuyin ddng lai gdn d td (2) vdi van td'c Vi 2 va gap nhau sau khoang thdi gian t = 15 phut = 0,25 gid, nghia la di hit doan dudng s = 20 km. Do dd : V, -) = -s = 20 km = 8o0^ k, m//h. '^ t 0,25 h Thay Vi 2 = 80 km/h vao trdn, ta ed : (1) Vl,2 = Vl,3+V2,3 = 8 0 120

- Khi hai d td chay cflng ehilu nhau thi ca ba van td'c Vj 3 ; Vi 2 ; V2 3 diu cung phuong ehilu. Do dd theo edng thflc cpng van tdc ta ed : Vl,3 = v'1,2 + ^2,3 suy r a : vj 2 = V13 - V2,3 (2) Thay v, -, = — = = 20 km/h vdo (2), ta lai cd : '^ t' 1 h vl,2 = Vi,3-V23 = 20 (3) Giai he hai phuong trinh (1) va (3), ta tim dupc van tdc cua hai d td : Vl T = 80 + 20 = ,5^0. km„/h 2 V2 3 = = 30 km/h Ghi chii : Cd thi giai bai todn nay bdng cdc phuong trinh chuyin dpng. Chpn true toa dp trflng vdi quy dao thdng, chpn bin A lam gde toa dp va ehilu duong hudng tfl A dd'n B, chpn thdi dilm xud't phat eua hai d td lam gd'c thdi gian. - Khi Jiai d td chay ngupc chieu thi phuong trinh ehuyin dpng cua 0 tP (1) di tfl A va cfla d td (2) di tfl B cd dang : Xi=Vit X2 = 20 - Vjt Khi hai d td gap nhau thi X2 = Xi, nghia Id : Vit = 20 - V2t . (4) 20 hay ''I \"^^2 = - 5 ^ ==80 - Khi hai d td chay cung chiiu thi phuong trinh chuyin dpng cfla d td (1) di tfl A va cua d td (2) di tfl B ed dang : Xi=Vit X2 = 20 + Vjt 121

Khi hai d td gap nhau thi Xj = Xi, nghia la : vit = 20 + V2t (5) 20 hay Vl - V2 = — = 20 Giai hd hai phucmg trinh (4) va (5), ta lai tim dupe kit qua tuong tu nhu each giai ndu d phdn trdn : Vi = 50 km/h va V2 = 30 km/h. 6.8. Gpi Vl 2 la vdn tdc cua ca nP (1) ddi vdi ddng chay (2), V23 la vdn td'c cua ddng chay (2) ddi vdi bd sdng (3) va Vi 3 la vdn td'c cua ca nd (1) ddi vdi bd sdng (3). a) Khi ca nd chay xudi ehilu ddng ehay thi cac vdn tdc Vj 2 va V2 3 cd cflng phuong chiiu, ndn theo cdng thflc cpng vdn tdc thi vdn tdc Vi 3 cua ca nd (1) ddi vdi bd sdng (3) ed gia tri bdng : Vl,3 = Vi,2 + V2,3 Thay Vi 3 = - = = 24 km/h va V2 3 = 6 km/h vao, ta suy ra dupc t 1,5 h gid tri van td'c Vi 2 cua ca nd ddi vdi ddng ehay bdng : Vi,2 = Vl 3 - V23 = 24 - 6 = 18 km/h b) Khi ea nd chay ngupe chiiu ddng chay thi cdc vdn tdc Vi 2 va V2 3 ngupc ehilu, ndn vdn tde Vj 3 cua ca nd ddi vdi bd sdng trong trudng hpp nay co gid tri bdng: v'1,3 =Vi,2-V2,3 Thay sd, ta tim dupe : v'l 3 = 1 8 - 6 = 1 2 k m / h Nhu vdy khoang thdi gian ngdn nhd't dl ca nd chay ngupc ddng chay tfl ben B trd vl dd'n bin A se bdng : t ' = ^ = ^ = 3h v;,3 12 Thdi gian ehay ngupe ddng chay ldn gdp ddi thdi gian chay xudi ddng chay. 122

6.9. a) Gpi Vl 2 la van td'c cua ca nd (1) ddi vdi ddng nudc (2) khi nudc dung ydn, V2 3 la van tdc cfla ddng nudc (2) ddi vdi bd sdng (3) va Vi 3 la van td'c cfla ea nd (1) ddi vdi bd sdng (3). Thdi gian ehay xudi ddng la ti va thdi gian chay ngupc ddng la tj. - Khi ca nP chay xuPi ddng tfl bd'n A vl bin B thi: Vl,3 = Vl,2+V2,3 TU A B s. . Thay Vi 3 = = - vao ta cd : '-' tl 2 f = 3 0 + V2,3 (1) - Khi ca nd chay ngupc ddng tfl bin B trd lai bin A thi : Vu =Vl,2-V2,3 r^ • AB s . Thay Vi 3 = = — vao ta co : 1 = 30 - V2,3 (2) Giai he phupng trinh (1) va (2), ta tim dupc khoang each gifla A va B : - + i = 60 => s = 72 km. 23 b) Tfl (1) ta suy ra vdn tdc cua ddng nudc ddi vdi bd sdng bdng : V23=- - 3 0 = — - 3 0 = 6km/h. ^•^ 2 2 6.10. Cach giai tuong tu bai tap tren. Gpi Vi 2 la van tdc cua ca nd (1) ddi vdi ddng nudc (2) khi nudc dung ydn, V2 3 la van td'c cua ddng nude (2) ddi vdi bd sdng (3) va Vi 3 la van td'c cua ca nd (1) ddi vdi bd sdng (3). Thdi gian chay xudi ddng la tj va thdi gian chay ngupc ddng la t2. - Khi ca nd chay xudi ddng tfl bin A ve bin B, ta cd : Vl,3 = Vl,2 + V 2 , 3 = 7 - 4 - Khi ca nd chay ngupc ddng tfl bin B trd lai bin A, ta cd : v'1,3 = V l , 2 - V 2 , 3 = - ^ to 123

suy ra: 1 _ S(t2-ti) ^2 J 2tit2 ^2,3 = - Vti Nlu ea nd bi tat may va trdi theo ddng thi van td'c cfla ca nd ddi vdi bd sdng dung bdng van tdc cua ddng nudc ddi vdi bd sdng, nghia la : Vi 3 = V23 Gpi t3 la thdi gian dl ea nd trdi xudi ddng tfl A dd'n B, ta cd : s t3= V2,3 Thay bilu thflc cfla V2 3 tim dupc d trdn, ta tim dupe : 2tit2 2.2.3 12 gid. t3 = 3-2 BAI TAP CUOI CHUONG I 1.1. B ; 1.2. D ; 1.3. B ; 1.4. D ; 1.5. A ; 1.6. C ; 1.7. D ; 1.8. C. 1.9. a) Phudng trinh chuyin dpng cua d td : x^ = 80t Phuong trinh chuyin dpng cfla xe may : Xg = 20 + 40t b) Hai xe gap nhau khi x^ = Xg. Tfl dd suy ra : - Thdi dilm hai xe gap nhau kl tfl khi x (km) I xud't phdt: 160 n^ 80t = 20 + 40t hay t = 20 = 0,5 gid =30 phut 120 80-40 80 - Vi tri hai xe gap nhau each A mpt M doan : x^ = 80.0,5 = 40 km. 40 c) Do thi toa dp cua hai xe cd dang nhu 0,5 1 1,5 t(h) tren hinh I.IG, trong dd dudng I bilu Hinh I.IG 2,0 didn chuyin ddng cfla d td va dudng II bilu diln chuyen dpng cfla xe may. Can cfl vao dd thi trdn hinh I.IG, ta thd'y hai dudng bilu diln I va II giao nhau tai dilm M ung vdi thdi diem hai xe gap nhau t = 0,5 gid = 30 phut d vi tri cd toa dp X = 40 km. Nhu vdy kit qua tim dupe trdn dd thi trung vdi kit qua tfnh toan Uong cau b). 124

1.10. Chpn thdi dilm d td di qua dilm A lam mde thdi gian. Vi d td ehuyin dpng thang nhanh ddn diu ndn gia tdc cua P td dupc tinh theo ePng thflc : a=^B-VA (1) Mat khae gia td'c a lai lidn hd vdi quang dudng di dupe s va cae vdn tde v^ va Vg theo ePng thflc : VB - VA = 2as a) Ta suy ra 2s = (vg + VA)t hay v^ = Tfnh bang sd : v^ = 2.20 12 = 8 m/s Thay sd vao (1) ta tfnh dupc gia tde cua d td : a = 1 2 - 8 = 2- m/.s2 b) Vi vdn td'c ddu VQ = 0, ndn quang dudng di dupe cfla P to kl tfl dilm khdi hdnh cho din dilm A tinh bdng : _ atA SA = Vi v^ = at^ , ndn suy ra _ atA _ i 2a SA = Tfnh bdng sd: SA = — = 16 m. ^ 2.2 1.11. a) Vdn td'c xe dap trude khi ham phanh : ^ - . - 12 000 10 . VQ = 12 km/h = = — m/s. ° 3 600 3 Ap dung cdng thflc giua vdn tdc, gia td'c va quang dudng di dupc v^ - VQ = 2as vdi V = 0 va s = 10 m ; ta dupc a = — « -0,55 m/s^. 2 Vdy, gia tdc cua xe dap la - 0,55 m/s . 125

b) Ap diing cdng thflc tfnh van td'c v = VQ + at ta dupc : t = —90 = ^6 s 15 vay, thdi gian ham phanh la 6 giay. 1.12. a) Gia su hdn bi chuyin dpng thdng nhanh ddn diu. Ta hay tim quy ludt bid'n ddi cfla nhung quang dudng di dupc lien tilp trong nhiing khoang thdi gian bdng nhau. Dat /i = AB ; /2 = BC ; /3 = CD ; /4 = DE. Gpi At la nhiing khoang thdi gian bdng nhau lien tie'p ma hdn bi chuyin dpng tren cdc doan dudng AB, BC, CD va DE. Gia sfl hdn bi xudt phat khdng vdn tdc ddu tfl dilm O va sau khoang thdi gian t nd lan de'n dilm A. 12 Gpi a la gia tPc cua hdn bi, ta cd OA = — at = s (1) O B = -a(t + At)^=s + A B (2) O C = -a(t + 2At)^=s + A B + B C (3) OD = -a(t + 3At)^=s + AB + BC + CD (4) 0E = -a(t + 4At)^=s + AB + BC + CD + DE (5) Ldn lupt lam cdc phep trfl ve vdi v l cdc phuong trinh trdn, ta cd (2) - (1): AB = atAt + -aAt^ = /, 2 (3) - (2): BC = atAt + -aAt^ = T 2 (4) - (3): CD = atAt + -aAt^ = l^ (5) - (4): DE = atAt + -aAt^ = U 126

Tiir cdc kit qua trdn, ta rut ra nhdn xet sau : /2-/1 =aAt^ ; /j-Zj = a A t ^ l^-l^ =aAt^ hay /2-/,=/g -/2 =/^ - / j = khdng ddi. Vdy, trong chuyen dpng thdng nhanh ddn diu, hidu nhflng quang dudng di dupe trong hai khoang thdi gian lien tid'p bdng nhau la mpt lupng khong ddi. Ap dung vdo bai toan nay (AB = 3 cm, BC = 4 cm, CD = 5 cm va DE = 6 cm) ta thd'y : BC - AB = CD - BC = DE - CD = 1 cm Vdy, chuyin dpng cua hdn bi la chuyen dpng thdng nhanh ddn deu. b) Tfl phip tfnh tren ta rflt ra cdng thflc tfnh gia tdc cfla hdn bi la l2-h —2 2 2 Vdi l2-l\\= 1 cm ; At = 0,5 s ; ta cd a = 4.10 m/s = 4 cm/s . 1.13. Gpi s la quang dudng roi cua gipt nudc mua tfl Iflc ddu dd'n dilm each mat ddt 100 m, t la thdi gian roi trdn quang dudng dd, ta cd : s =|gt' (1) Mat khae, quang dudng roi tfl lfle ddu dd'n mat dd't la s + 100 va thdi gian roi trdn quang dudng dd la t + 1 giay. Ta ed : s + 100 = ^ g ( t + l)2 (2) Tfl hai phuong trinh (1) va (2) ta rflt ra : t = —-0,5«9,70s g s a 461 m vay, dp cao ban ddu cua gipt nudc mua lfle bdt ddu roi la s + 100 = 561 m. 127

Chxtangii DONG Lire HOC CHAT DIEM BAI 9 9.1. C. 9.2. D. 9.3. B. 9.4. C. Gpi y : Xem hinh 9.1G. Luc Fi cd dp ldn F, luc F2 cd dp ldn 2F. 9.5. 49 N ; 69 N. Hinh 9.1G Hpp luc P' cua hai luc Fi va F2 cdn bdng vdi trpng luc P efla vat. Tfl hinh 9.2G tacd: P' = P = mg = 5,0.9,8 = 49 N. F = -P — = tan45° = 1 ^ R = P' = 49 N Fl i : = cos45«=^ F2 2 ^ F2 = P'V2 = 49.1,41 = 69 N. 9.6. 242 N. Dilm O coi la chdt dilm dflng cdn bang dudi tdc dung efla ba luc : trpng luc P va hai luc keo Fi va F2 cua hai nua ddy cdp (H.9.3G). B ^F B' o p Hinh 9.3G 128

Tfl hai tam giac lue ddng dang ta cd : Fi^ OB 2Fi OB F \" AB ^ P ~ AB 2 J P V A B ^ + O A ^ ^ 607025776 ^ ^41,86 . 2 4 2 N. 2.AB 2.0,5 BAI 10 10.1.1-d ; 2-b ; 3-a ; 4-c. 10.2. B. 10.3. D. 10.4. Khdng. Vdt cd the chiu nhilu luc tac dung, nhung cac lue nay la edc luc cdn bdng. 10.5. Sai. Do cd qudn tfnh, tui sdch bao toan vdn tdc khi xe dtmg lai dpt ngpt, nen bay vl phfa ddu xe. 10.6. Do cd ma sdt. 10.7. Do khdng loai bd dupc trpng luc va luc ma sdt. 10.8. Xe mdy se ddm vdo phfa sau xe tai. - Do phan xa cfla ngudi lai xe may la khdng tfle thdi ma edn ed mpt khoang thdi gian du rd't ngdn dl nhan ra xe tai da dflng va dn chdn vao phanh. - Do xe cd quan tfnh, ndn du da chiu luc ham cung khdng thi diing lai ngay ma cdn cd thdi gian dl dflng hdn. Trong hai khoang thdi gian ndu tren, xe may kip di hit khoang each giua hai xe va ddm vao xe tai. 10.9. Khi xe dang chay nhanh ma phanh gdp, ddy an todn giu cho ngudi khdng bi lao ra khdi ghi vl phia trudc. Khi xe dpt ngpt tang tdc, cdi tua ddu giu cho ddu khdi ngdt manh vl phia sau, trdnh bi dau cd. 9.BTVATLilO-A 129

10.10. 1-^ c ; 2 ^ d ; 3-> a ; 4-> d ; 5-> b. 10.11. B; 10.12. C. 10.13. D; 10.14. C. 10.15. B; 10.16. D. 10.17. B; 10.18. C. 10.19. D; 10.20. A. 10.21. Ngudi cheo thuyin dflng mai cheo tac dung vao nudc mdt luc hudng vl phfa sau. Nudc tdc dung lai mdi cheo mdt luc hudng vl phia trudc lam thuyin ehuyin dpng. Khi cdnh quat cua may bay quay, nd ddy khdng khf vl phfa sau. Khdng khf ddy lai ednh quat vl phfa trudc lam may bay chuyin ddng. 10.22. 3 kg. Chpn ehilu duong la ehilu chuyen dpng cua vdt 1. C TT ^Vi AVj F21 = -F12 => mjai = -m2a2 ^ m^—j- = -mj—f- At At mi[(-l)-5] = -m2(2-0) m2 = 3mi = 3 kg. BAI 11 11.1. B. 9BTVATLI10.B 11.2. D. 11.3. 54R. 130

Gidi Gpi X la khoang each tfl dilm phai tim dd'n tdm Trai Dd't, Mi va M2 ldn lupt la khdi lupng cua Trdi Ddt va cua Mat Trang, R la bdn kfnh Trdi Dd't va m la khdi lupng cua con tau vu tru. GMjm _ GM2m ^ _ _ 1_ hay X = 54R. x^ \" ( 6 0 R - x ) ^ x\"(60R-x) 11.4. 9,79 m/s^ 4,35 m/s^. GM GM g= g= R^ (R + hr Suy ra R a) g =g R + h, h = 3 200 m = 3,2 km g' = (9,8) ^66440030,2 ^^ = 9,79 m/s^ b) h = 3 200 km 6400^ g' = (9,8) 9600 = 4,35 m/s^ 11.5. 735 N ; 127,5 N ; 652,5 N ; 0 N. a) P = mg = 75.9,8 = 735 N. b) P = mg = 75.1,7 = 127,5 N. c) P = mg = 75.8,7 = 652,5 N. d) P = 0. BAI 12 12.1. A. 12.2. B. 12.3. 27,5 cm. F/, = P = mg F/x = k(/-/o) 131

Suyra: ^2 \" ' o ^ ni2 /l - IQ mi Ij - 2 5 , 0 _ 100 0,5 20 I2 = 27,5cm. 12.4. 7,5 N. Fn,ax-k(/,ax-W = 7 5 ( 3 0 - 2 0 ) . 1 0 - ^= 7,51 12.5. 14 cm ; 60 N/m. Fi = k ( / i - / o ) ; F2 = k ( / 2 - / o ) ; F2 h-lo ^ 4,2 2 1 - / 0 Fl / 1 - / 0 1,8 1 7 - / 0 ^ 1,8(21-/Q) = 4,2(17-/Q) /o= 14 cm k = ^1 = ^'^0 = 60 N/m. ^1-^0 3.10~2 12.6.30 cm; 100 N/m. F,x = P=^ k ( / - / o ) = mg ^1-^0 _ mi Thav\"d- ^^~^0 _ 100 /2 - /Q mi + m2 32 - IQ 200 k = \"^IS = ^ ' 1 - 1 ? =100 N/m. ' 1 - ^ 0 1.10~2 12.7. 294 N/m; 2,4 N. F/, = k(/-/o) = P = ^ k = - ^ = — ^ 294 N/m. ' l - ^ 17.10\"\"* i - ^ = ^=>P2.Pl / 35-27 44-27 h-h P2 7l-^ P2 = 2,35 « 2,4 N. 132

12.8. 245 N/m ; 0,38 kg. F/x = k ( / - / Q ) = mg. ^ k = J ^ = 0,50.9,8 , 2 4 5 N/m. ^ 1 - ^ (7,0-5,0).10~2 mi _ /i -/o _ mi(/2 -/o) _ 0,50.1,5 m2 / 2 - / 0 /1-/0 2,0 = 0,375 kg « 0,38 kg. 12.9. 56 N/m ; 2,5 N. a) Vi F ti Id thudn vdi A/. b) k = — = — ^ = 55,5 « 56 N/m. A/ 9.10\"^ c) F = ^'^ ^ ^ ' ^ = 2,45 « 2,5 N. BAI 13 13.1. A. ' 13.2. C. 13.3. C. 13.4. 2,1m. Chpn ehilu chuyin dpng la ehilu dudng. -F^s = ma ^ -umg = ma => a = -|Lig v2-v^=2asz^s =i = -&5)L„2,lm. 2ng 2.0,30.9,8 13.5. a) Dl tang ma sdt nghi. b) Mat vai da la thudng nhan, ma sat giam, bui khd bam. c) Khi can cudc dm, cdc thd gd phdng ldn, ma sdt tang ldn dd cdm hdn. 133

13.6. Vi luc ma sat nghi cdn bdng vdi luc keo. 13.7. 0,56 m/s^. Chpn ehieu cfla lue F lam chiiu duong. F^5 = i^mg = 0,35.55.9,8 = 188,65 N a = F - F „5„^ 220-189 « 0^,,,56 m/,s2. m 55 13.8. 440 N; 0,056. a) Luc ma sat nghi da gdy ra gia td'c cho d td ^Av^ Fmsnmax = ma = m ^Aty* 800.20 = 444,4 « 440 N 36 b) ' msn max ma a 20 mg g 36.9,8 = 0,056. BAI 14 14.1. 5,6 km/s ; 240 phflt; 1500 N. a)Fhd = Fht 2 GM GMm mv V= 4R2 2R 2R Mat khde, trdn mat dd't ta cd : P = mg = —GM—m ^ g = GM = 9,8 mis' R^ R2 Suyra V = JR—g^ = J64.10^.9,—8 =^5^^6^00 m, /s V2 V 2 V = 5,6 km/s b) T 47iR _ 4.3,14.64.10^ = 14354,29 s « 240 ph. V \" 5600 134

c) P^, . i ^ = 600.(5600)2 ^ 6.(5600)2 ^ ^ ^^^ ^ ^ ^^^ ^ 2R 2.6400000 2.64000 Fhd « 1 500 N. 14.2. 6,00.10^^ kg. Gpi M va m ldn lupt la khdi lupng cua Trai Dd't va cua Mat Trang, r la bdn kfnh quy dao efla Mat Trang. „„ GMm 2c A/r ^^r^ Fjid = Fht => —^— = meo r. Suy ra : M = -—- r^ G 271 Thay co = — vdo ta dupc : 4nV M= T^G 4.3,14 2.3—,8T4^102^ -j- (1 nga. y = 86 4c0,.0.rs^). (27,32)2.(864)2.10 .6,67.10\"\" 2233.1024 746,4.746,5.10^.6,67.10\"\" M « 6,00.10^^ kg. 14.3. 9 2 0 N ; h = 153 km. a) Fht = P = 920 N b)Fht= ma)2r=920N Suy ra : r = 920.T2 920.(5,3)2.10^ ^= m.4n 100.4.(3,14)2 = 65,53.10V = 6 553km h = r - R = 6 553 - 6 400 = 153 km. 14.4. 5 kg. 14.5. 8,88 N. 135

14.6. 1,19 m/s (xem hinh 14. IG). F^t = mgtana 22 Fht = mv mv r /sina .2 'ht r» Suy ra : mv = mg tana mg /sina Hinh 14.1G FT—. /9,8.0,5.0,5 V = ^Jgl sina tana = J 7= = 1,189 « 1,19 m/s. 14.7. 19 000 N. Fht = mg - N mv N=m _v^l f 15^ = 2500 9,8 V 100 V 'J = 18 875 « 19 000 N BAI 15 15.1. C. 15.2. D. 15.3. B. 15.4. 42 m/s. h=1 2 2.90 = 4,2 s 9,8 L.ax=VQt=^VQ=-f^ = 42m/s. 136

15.5. 3,2 s ; 36 m/s. a) y => t = 2h _ &^ = 3,19 » 3.2 s. -'-\\^' ' 9,8 b) V = ^/v?+v2 =^vg+(gt)2 = 36,06 ^ 36 m/s. 15.6. 10 s; 1 500 m. , 2h 2.490 ,_ a) t = J — = J = 10 s V g V 9,8 b) Gpi VQ la tdc dp cua gdi hang khi rdi khdi may bay. Ta cd: Lmax = VQt = 150.10 = 1500 m. c) Quy dao parabol. BAITAPCUOICHaONGII 11.1. A. 11.2. B. 11.3. a) BQii budc ldn bdc cdu thang, chdn ngudi da tac dung vao bdc mpt luc hudng xudng. Bdc cdu thang tac dung lai chdn ngudi mpt phan luc hudng ldn, luc nay ndng ngudi ldn bdc trdn. b) Qua bdng tdc dung vao lung dfla tre mpt luc. Lung dfla tre tdc dung lai qua bdng mpt phan lue lam qua bdng bdt trd lai. 11.4. Cap \"luc va phan luc\" : a va b. Cap luc cdn bdng : c va d. 11.5. Chpn chiiu duong la ehilu chuyen dpng luc ddu cua qua bdng. Luc ma gay dap vao qua bdng la : mAv „ ^ f (-20) - (30)1 F = ma = = 0,2^^ '^ = -400 N. At 0,025 Luc ma bdng tac dung vao gay la : F' = - F = 400N. F' > 0 => luc F' hudng theo chiiu ehuyin dpng ban ddu cua qua bdng. 137

11.6. 250 N. Theo dinh ludt III Niu-ton, d ca hai trudng hpp, luc cua dpi A keo day va luc cua dpi B keo ddy diu la cap \"luc va phan luc\", do dd diu ed dp ldn bdng nhau, tflc la bdng 250 N. a) Hai dpi hod la vi hai dpi cung dap chdn vao mat dd't vdi mdt luc cd dp ldn bdng nhau. Theo dinh ludt III Niu-ton, phan luc ma mat dd't tac dung vao hai dpi cung cd dp ldn bdng nhau. Nd'u xet ridng tiing dpi, thi luc keo cua ddi phuong va phan luc eua mat dat tac dung vao mdi dpi cdn bdng nhau lam mdi dpi dung ydn (H.II. IGa). b) Dpi A thdng la vi dpi A dap chdn vdo mat dd't vdi mpt luc ldn hon. Theo dinh ludt III, mat dd't tdc dung lai dpi A mdt luc ldn hon luc ma dpi B keo dpi A, lam dpi A thu gia td'c va chuyin dpng keo theo dpi B chuyin dpng vl phfa minh (H.II. 1Gb). 'BA •BA -J* w/////^777?/7mim'A??^ - -S W/>///'/^/^w>n<i'i'i',iWt'i^ -• f^DSt-A 'oft-A 'A-Bat 0\\-DSt a) b) Hinh II.IG 1.7. 6°40' ; 16 cm. a) CO = 27if = ^ 1 ^ ^ = 344 rad/s 60 99 tana _ 'PYA_ _ mcc) r _ (0 / sina P mg g g _ 9,8 cosa = 0)2/ (3,14)2.1,00 0,9940 a » 6°40' Hinh n.2G b) V = cor = (o/sina = 3,14.1,00.0,1167 V = 0,366 m/s ^ 2h 2^^ V g V 9,8 s = vt = 0,366.0,452 = 0,165 m = 16,5 « 16 cm. 138

11.8. 3,0 s ; 760 m ; 30 m/s. ^ . 2h 2.45 = 3,03« 3 s. '9,8 b) Lmax = Vot = 250.3,03 = 757,5 « 760 m. c) Vy = gt = 9,8.3,03 = 29,7 « 30 m/s. Chttangiii CAN BANG j> VA CHUYEN DONG CUA VAT RAN BAI 17 17.1. 25 N; 43 N. P'=-p Vdt chiu tdc dung eua ba luc cdn bdng : trpng lue P , phan luc vudng gdc N eua mat phdng nghidng va luc cangf cfla ddy (H. 17. IG). Tfl tam gidc luc, ta ed : - = sin30° =0,5 P T = 0,5.5.10 = 25 N N_ = cos30° Hinh 17.IG P\" • ^ N = Peos30° R N = 5.10.—= 43N 2 Ap luc N' cfla vdt vao mat phang nghidng la luc true ddi vdi phan luc N cua mat phdng nghidng ldn vat. Suy ra N' = 43 N. 139

17.2. 40 N ; 56 N. Dilm C dung can bdng (H.17.2aG), ndn Ti* Tl = P = 40 N Thanh chdng dung can bdng (H.17.2bG), ba luc fj, % va Q ddng quy d B. Tfl P• b) tam giac lire, ta cd : a) Hinh 17.2G Q = Tl = P = 40 N T2= T1V2 =56,4«56N. Chu y : Do tudng khdng cd ma sat ndn xfch phai cd ma sat mdi gifl dupc thanh chd'ng, vi vay T2 phai ldn hon Ti. 17.3. ION; 17 N. Thanh AB chiu ba luc can bdng la P, Nl va N2. Vi mat phang nghidng khdng ma sat ndn hai phan luc Ni va N2 vudng gdc vdi cdc mat phdng ^^-^^ jy^Q nghieng. Ta trupt cdc vectd lue trdn gid eua chung dd'n dilm ddng quy C (H.17.3G). Tfl tam gidc luc, ta dupc : Nl =Psin30° = 2 0 . - = 10N 2 /3 N2 = Pcos30° = 2 0 . — = 17,3 « 17 N. 2 Theo dinh ludt III Niu-tcm thi dp luc cua thanh ldn mat phang nghieng cd dp ldn bang phan luc cua mat phang nghidng ldn thanh. 140

17.4. ~ . Hf B V3 Hinh 17.4G Gpi Fg la hpp luc cua lue cang f va phan luc Ng eua san. Ta ed he ba luc can bdng la P, N ^ va Fg. Ba luc nay ddng quy tai C(H.17.4G). /T Vi OA = CH = O B ^ ^ ndn tam giac OCB la tam giae diu. Tfl tam gidc luc ta cd : T = NA=Ptan30°=^. V3 BAI 18 18.1. 40 N ; 500 N/m. a) Mp^^ = Mp F/x.OC = F.OA F/^ = 2F = 40 N. b ) k = ^ = —=500N/m. A/ 0,08 18.2. 6 N. 18.3. 86,5 N ; 100 N. MF = MP a)F/ = P-^cos30° 2 F = ^ = ^ ^ = 86,5N. 44 b)F/cos30° = P^cos30° F = - =100N. 2 141

18.4. 40 N. Coi mep ban la true quay 0 Mp == MF pi 44 p = F = 40 N. 18.5. ION. Xem hinhl 8. IG. Thanh cd true quay cd dinh O, chiu tae dung cfla ba luc P, T va Q. Ap dung quy tde momen luc, ta dupc : M^O) = M^P^ T.OH = P.OG T-.OA = P-OA -TTi 22 i '/// T = P = mg= 1,0.10= ION. i 18.6. a)400N;b)350N. I a) Xet momen lue ddi vdi true quay O : ' MT = MT • T2/sina = T^l fa.- * • T2 = 4 _ 200 400 N. h/- 0,5 sma F b) Hpp lue F cfla hai luc Tj va T2 phai H/ \"-. hudng dpe theo thanh vao O. N, F = Tjcosa = 400^/3 = 346 = 350 N. Hinh 18.2.G BAI 19 19.1. B. 19.2. a) 100 N ; b) 25 N ; c) 150 N, 75 N. a) - = — = 2 ;^ F = 2P = 100 N. P 30 142

KX F 30 1 1 b) - = — = - ^ F = -!-P = 25 N. P 60 2 2 flSO N (trudng hpp a) c) Ap lue bdng F + P = 75 N (trudng hpp b) 19.3. 107 N ; 193 N. Ta phdn tich trpng luc Pj cua true thanh hai luc thanh phdn tac dung len hai d truc A va B : PiA = PiB = f = 50 N. Lam tuong tu vdi trpng luc P2 cua banh da : P'^o2AA +P'^?^R28 = Po = 2 0 0 N ^2A 04 0,4 •^2B 1 Giai hd ta dupc P2A = 57 N va Pjg = 143 N. Vdy, dp luc ldn d true A la Pi^ + Pj^ = 107 N. Ap luc len d tnic B la Pig + P2B = 193 N. 19.4. 1 800 N.m ; 2 400 N. • ^1 a) M = P/ = 600.3,0 = 1 800 N.m. b) Momen eua luc F2 cfla cpc dd sau ddi vdi cpc dd trudc phai cdn bdng -1 m- 3m vdi momen eua trpng lue cua ngudi. Do dd, luc F2 phai hudng xud'ng ^ ,, (H.19.1G). Mp = F2d2 = 1 800 N.m Hinh 19.10 => F2 = 1 800 N. Hpp luc cua F2va P can bdng vdi luc F^ Fi=F2+P = 2400N. 143

BAI 20 20.1. Dien tich tilp xue cfla thude vdi ban la didn tfch mat chdn dl. Khi thudc nhd ddn ra khdi mep ban thi didn tich mat chan d l bi giam ddn. Thudc bat ddu roi khi trpng tdm roi vao mep mat chdn de cfla ban, cung la mep ban. 20.2. 45°. '////////////////}//// Coi khdi ldp phucmg la mpt vat cd mat chan dl. Gdc nghidng a cue Hinh 20.1G dai khi trpng luc cd gid di qua mep A cua mat chdn dl. Tfl dd suy ra «m = 45°(H.20.1G). 20.3. 29°. Xem AB la mat ehdn d l (H.20.2G). tanttj II . 1'2 = 0,5454 \"2,2 ffi ffi ttm = 28,6°. 20.4. 3/ 4• Gia sfl vidn gaeh 2 khdng bi dd thi vidn gaeh 3 ehi dupe phep nhd ra ,/ Hinh 20.2G khoi vidn gaeh 2 cue dai la — m112 •• 2 1 (H.20.3G). '777777777777*77771717477Z Dung quy tdc hpp luc song song cflng ehilu ta thd'y trpng tdm G cua hai vidn gaeh 3 va 2 d cdch mep phai cfla vidn gaeh 2 mpt doan —/. Do dd vidn gaeh 2 chi dupe phep nhd ra khdi vidn gaeh 1 dudi cflng mdt doan —/. Hinh 20.30 •4 Vdy vidn gaeh trdn cflng chi dupc phep nhd ra khdi mep phai cua vidn gaeh dudi eung mdt doan la L L-E. 2 4 \" 4\" 144

BAI 21 21.1. B. 21.2. 0,375 m/s ; hudng trflng vdi hudng eua luc, tfle la ngupe vdi hudng cua chuyin dpng. 21.3. a) 2s ; b) - . 4 Chpn ehilu duong la ehilu chuyin dpng. v^ - Vn = 2as a=^ = - y* 2s m Suy ra : s = mvp 2F a) s, = 2mv^ = 2s. 2F b) S2 = mv'0 _ 2F.4 4 21.4. 3,34 N; 2,94 N. Hinh 21.1G ve cac luc tac dung len vat. t4 N ,F Fms = MtN Phuong trinh ehuyin dpng cua vat theo cac v/z/Ty/// V////////////A phuong Ox, Oy cd dang : p Ox : F - iJ.tN = ma Hinh 2 I.IG Oy: N-mg = 0 Suy ra F = m(a + n,g) = 1,0.(0,40 + 0,30.9,8) F = 3,34 N. b) F = F„s = W^B = 0,30.1,0.9,8 = 2,94 N. 10.BTVATU'10-A 145

21.5. 0,26. y* Hinh 21.2G ve cdc luc tdc dung ldn mg vdt. Hinh 21.2G Phuong trinh ehuyin dpng efla vat theo cdc phuong Ox, Oy cd dang : Ox: Fcos30°-F^s = ma (1) Oy: N + Fsin30° - mg = 0 (2) F„,s = HtN (3) Tfl (1), (2) va (3) ta tim dupc : N = mg - Fsin30° Fcos30° - |J.t(mg - Fsin30°) = ma Feos30° - m a = 120.0,866-32.1,2 = 0,256 =>|a,«0,26. Suy ra it, = ' mg-Fsin30° 32.9,8-120.0,5 21.6.30° ;l,3m. a)ffinh21.3Ga Ox : Psina = ma (1) Oy: N-Pcosa = 0 (2) Mat khae, theo bai ra : a = 2s (3) Tfl (1), (2), (3) suyra V77777777777777777777; sina = —a = —2s - = 2.2,45 = 0,5 a) g gt^ 9,8.1 ^ a = 30°. b) Hinh 21.3Gb. mgsina - JJ-JN = ma (4) N - mgcosa = 0 (5) Hinh 21.3G s = 1- a t2 b) 2 (6) 146 lO.BTVATUlO-B

Tfl (4) va (5) => a = g(sina + Mt^osa) Hinh 21.4G a = 9,8(0,5 - 0,27.0,866) = 2,606 = 2,6 m/s^ Tfl(6): s =-.2,6.1 = 1,3 m. 2 21.7. 1039N(H.21.4G). Fi2 = 2Fieos30° F3 = F12 BAI 22 22.1. D. 22.2. Khdng thay ddi. 22.3. Xem hinh 22.IG. a) 1,6 N.m ; b) 0,80 N.m ; e) 1,38 N.m. b) c) Hinh 22.1G Hinh 111.10 BAI TAP CUOI CHaONG 147 1.1. C. 1.2. 1,57 m/s2; 63,5° theo hudng Ddng-Bdc. Tfl hinh m.lG, tacd: F = ^/pf + F | = V3802 + 1902 F - 425 N

tana = ^ = 2 a = 63,5° F 425 . „ ,2 '777777777. a = — = = 1,57 m/s . Hinh 111.20 m 270 .3. 0,29R. Con lan vupt qua dupe bdc them nd'u momen eua luc F ddi vdi true quay A ldn hon hodc bdng momen cua trpng ltic P (H.in.2G). F(R - h) > PVR^ - (R - h)^ F(R - h^) = pVR2-(R-h„)2 2h^ - 4Rh^ + R2 ^ 0 Vi ehi ld'y nghiem 0 < h < R ndn ta dupc h^^^ - 0,29R. 1.4. Ta phdn tich luc Pi thanh hai luc tac dung ldn hai cpt: P i i - P i 2 = f = ^ = 5000N. Lam tuong tu vdi luc P2 : 2 r•-^i2i1 + P-2io2 — P' mg tp. 2 L2I Hinh 111.30 3 f'22 Suyra: P 2 i = i ^ = H M = 1250N ^^8 8 Pr2„2 = ^ = 3 750 N. 8 Ap lue len cpt 1 la : Fj = Pn + P21 = 6 250 N Ap luc ldn cpt 2 la : F2 = P12 + P22 = 8 750 N. 148